a lot contains 20 articles.d probability that d lot contains xactly 2 defective articles is 0.4;and d probability dat it contains exactly 3 defective articles is 0.6.articles r drawn from d lot at random 1 by 1 without replacement until all d defective r found.Wat is d probability dat d testing procedure stops at d twelfth testing?

Hi,

Clearly there are two cases, either the lot as 2 deffective articles or 3 as the sum of probabilities is 1.

Now case 1 : When the lot has 2 defective articles. So we are going to draw one defective article between deaw no 1 to 11^th and will draw the second defective article at 12^th draw as this will mark the end of our search for this case.

So the probability will be 18/20.17/19 …...9/11.(2/10).1/9 (total 12 draws in which we have taken perfect articles till 10^th draw, on 11^th we took one defective and on twelth we took another defective)

Now we need to arrange one defective draw in 10 perfect article draws which will be : (11!)/(10!.1!)

Multiply this with the above series, this is result for first case,

Similary for the second case we have : 17/20.16/19.......9/12 . 3/11 . 2/10. 1/9 (i.e. 9 perfect article draws and there on 3 defective draws, now we need to keep the twelth draw at its place and have to permute rest 11 draws in which 9 are perfect draws and 2 are defective so : (11!)/(9!.2!) Multiply this with the second case series.

Add the results of two cases, which will be ur answer. ie:
18/20.17/19 …...9/11.(2/10).1/9[(11!)/(10!.1!)] + 17/20.16/19.......9/12 . 3/11 . 2/10. 1/9[ (11!)/(9!.2!)]